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Chapter One
Section 1.
For C "Þ& , the slopes are negative , and hence the solutions decrease. For C "Þ&, the slopes are positive , and hence the solutions increase. The equilibrium solution appears to be C >a b œ "Þ&, to which all other solutions converge.
For C "Þ& , the slopes are:9=3 tive , and hence the solutions increase. For C "Þ& , the slopes are negative , and hence the solutions decrease. All solutions appear to diverge away from the equilibrium solution C >a b œ "Þ&.
For C "Î# , the slopes are:9=3 tive , and hence the solutions increase. For C "Î# , the slopes are negative , and hence the solutions decrease. All solutions diverge away from
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the equilibrium solution C >a b œ "Î#.
For C # , the slopes are:9=3 tive , and hence the solutions increase. For C #, the slopes are negative , and hence the solutions decrease. All solutions diverge away from the equilibrium solution C >a b œ #.
For all solutions to approach the equilibrium solution C >a b œ #Î$, we must have C w^ ! for C #Î$ , and C w! for C #Î$. The required rates are satisfied by the differential equation C wœ # $C.
For solutions other than C >a b œ # to diverge from C œ # , C >a bmust be an increasing function for C # , and a decreasing function for C #. The simplest differential equation whose solutions satisfy these criteria is C wœ C #.
For solutions other than C >a b œ "Î$ to diverge from C œ "Î$ , we must haveC w! for C "Î$ , and C w! for C "Î$. The required rates are satisfied by the differential equation C wœ $C ".
Note that C w^ œ! for C œ! and C œ &. The two equilibrium solutions are C >a b œ !and C >a b œ &. Based on the direction field, C w! for C &; thus solutions with initial values greater than & diverge from the solution C >a b œ &. For! C &, the slopes are negative , and hence solutions with initial values between! and &all decrease toward the
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7 œ 71 @
or equivalently,
.@ .> 7
œ 1 @ Þ
a b, After a long time, .@.> ¸! Þ Hence the object attains a terminal velocity given by
@ œ Þ
_ Ê^
a b- Using the relation # @ (^) _#^ œ 71 , the required drag coefficient is#œ !Þ!%!) 51Î=/- Þ
a b.
All solutions appear to approach a linear asymptote a A3>2 =69:/ /;?+6 >9 "b. It is easy to verify that C >a b œ > $is a solution.
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All solutions approach the equilibrium solution C >a b œ! Þ
All solutions appear to diverge from the sinusoid C >a b œ (^) È$# =38Ð> (^1) %Ñ ",
which is also a solution corresponding to the initial value C !a b œ &Î#.
All solutions appear to converge to C >a b œ !. First, the rate of change is small. The slopes eventually increase very rapidly in magnitude.
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Section 1.
1 a b+ The differential equation can be rewritten as
.C & C
œ .> Þ
Integrating both sides of this equation results in 68 & C œ > -k k (^) ", or equivalently, & C œ - / >^. Applying the initial condition C !a b œ C!results in the specification of the constant as - œ & C (^)!. Hence the solution isC >a b œ & aC & / (^)! b >Þ
All solutions appear to converge to the equilibrium solution C >a b œ & Þ
1 a b- ÞRewrite the differential equation as
.C "! #C
œ .> Þ
Integrating both sides of this equation results in "# 68 "! #C œ > -k k ", or equivalently, & C œ - / #>^. Applying the initial condition C !a b œ C!results in the specification of the constant as - œ & C (^)!. Hence the solution isC >a b œ & aC & / (^)! b #>Þ
All solutions appear to converge to the equilibrium solution C >a b œ &, but at a faster rate than in Problem 1a Þ
2 a b+ Þ The differential equation can be rewritten as
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.C
C &
œ .> Þ
Integrating both sides of this equation results in 68 C & œ > -k k (^) ", or equivalently, C & œ - / >^. Applying the initial condition C !a b œ C!results in the specification of the constant as - œ C &!. Hence the solution isC >a b œ & aC & / Þ (^)! b >
All solutions appear to diverge from the equilibrium solution C >a b œ &.
2 a b, Þ Rewrite the differential equation as
.C #C &
œ .> Þ
Integrating both sides of this equation results in "# 68 #C & œ > -k k ", or equivalently, #C & œ - / #>^. Applying the initial condition C !a b œ C!results in the specification of the constant as - œ #C &!. Hence the solution isC >a b œ #Þ& aC #Þ& / (^)! b #>Þ
All solutions appear to diverge from the equilibrium solution C >a b œ #Þ&.
2 a b-. The differential equation can be rewritten as
.C #C "!
œ .> Þ
Integrating both sides of this equation results in "# 68 #C "! œ > -k k ", or equivalently, C & œ - / #>^. Applying the initial condition C !a b œ C!results in the specification of the constant as - œ C &!. Hence the solution isC >a b œ & aC & / (^)! b #>Þ
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+C ! œ + C 5" a (^) " b ,.
That is, +5 , œ! , and hence 5 œ ,Î+. a b-. The general solution of the differential equation is C >a b œ - / +> ,Î+ Þ This is exactly the form given by Eq. a "(b in the text. Invoking an initial condition C !a b œ C (^) !, the solution may also be expressed as C >a b œ ,Î+ aC ,Î+ / (^)! b +>Þ
6 a b+. The general solution is : >a b œ *!! - / >^ Î#, that is,^ : >a b œ *!! a : *!! /! b>Î#. With : (^)! œ )&! , the specific solution becomes : > a b œ *!! &!/>Î#. This solution is a decreasing exponential, and hence the time of extinction is equal to the number of months it takes, say > 0 , for the population to reach zero. Solving *!! &!/ >^0 Î#œ !, we find that > 0 œ # 68 *!!Î&!a b œ &Þ() months. a b, The solution, : >a b œ *!! a : *!! /! b >Î#^ , is a decreasing exponential as long as : (^)! *!!. Hence *!! a: *!! / (^)! b >^0 Î#œ !has only one root, given by
> œ # 68 Þ
a b-. The answer in part a b, is a general equation relating time of extinction to the value of the initial population. Setting > 0 œ "# months , the equation may be written as
*!! *!! :
œ / !
which has solution : (^)! œ )(Þ('". Since :!is the initial population, the appropriate answer is : (^)! œ )*) mice.
7 a b+. The general solution is : >a b œ : /! <>. Based on the discussion in the text, time >is measured in months. Assuming " month œ $! days , the hypothesis can be expressed as : /! <†"^ œ #: (^) !. Solving for the rate constant, < œ 68 #a b , with units of per month.
a b,. R days œ R Î$! months. The hypothesis is stated mathematically as : /!
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written as .@Î.> œ 1 , with solution @ >a b œ 1> @ Þ! The object is released with an initial velocity @ (^) !.
a b,. Suppose that the object is released from a height of 2 units above the ground. Using the fact that @ œ .BÎ.> , in which B is the downward displacement of the object, we obtain the differential equation for the displacement as .BÎ.> œ 1> @ Þ! With the origin placed at the point of release, direct integration results in B >a b œ 1> Î# @ >#^!. Based on the chosen coordinate system, the object reaches the ground when B >a b œ 2. Let > œ Xbe the time that it takes the object to reach the ground. Then 1X Î# @ X œ 2#^!. Using the quadratic formula to solve for X,
X œ Þ
! È!
The positive answer corresponds to the time it takes for the object to fall to the ground. The negative answer represents a previous instant at which the object could have been launched upward a with the same impact speed b, only to ultimately fall downward with speed @ (^) !, from a height of 2 units above the ground.
a b-. The impact speed is calculated by substituting > œ X into @ >a b in part a b+ Þ That is, @ Xa b œ È@ #12^!.
10 a +, b b. The general solution of the differential equation is U >a b œ - / <>Þ Given that U !a b œ "!! mg , the value of the constant is given by - œ "!!. Hence the amount of thorium-234 present at any time is given by U >a b œ "!! /<>. Furthermore, based on the hypothesis, setting > œ " results in )#Þ!% œ "!! / <Þ Solving for the rate constant, we find that < œ 68 )#Þ!%Î"!!a b œ Þ"(' /week or < œ Þ!#)#) /day.
a b-. Let Xbe the time that it takes the isotope to decay to one-half of its original amount. From part a b+ , it follows that &! œ "!! / œ < U is U >a b œ U /! <>, in which U (^)! œ U !a b is the initial amount of the substance. Let 7 be the time that it takes the substance to decay to one-half of its original amount , U (^)!. Setting > œ 7 in the solution, we have !Þ& U (^)! œ U /! <^7. Taking the natural logarithm of both sides, it follows that < 7 œ 68 !Þ&a b or< 7 œ 68 # Þ
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a b/. Letting >be the amount of time after the source is removed, we obtain the equation "! œ #((Þ(( / ^ !Þ!!!$>^ Þ Taking the natural logarithm of both sides, !Þ!!!$ > œ œ 68 "!Î#((Þ((a b or > œ ##ß ((' hours œ #Þ' years.
a 0 b
15 a b+. It is assumed that dye is no longer entering the pool. In fact, the rate at which the dye leaves the pool is#!! † ; > Î'!!!!c a b d kg/min œ #!! '!Î"!!!a bc; > Î'! a b d gm per hour . Hence the equation that governs the amount of dye in the pool is
.; .>
œ !Þ# ; a gm/hr b_._
The initial amount of dye in the pool is ; !a b œ &!!! grams.
a b,. The solution of the governing differential equation, with the specified initial value, is ; >a b œ &!!! / !Þ# >Þ
a b-. The amount of dye in the pool after four hours is obtained by setting > œ %. That is, ; %a b œ &!!! / !Þ)^ œ ##%'Þ'% grams. Since size of the pool is '!ß !!! gallons , the concentration of the dye is !Þ!$(% grams/gallon.
a b.. Let Xbe the time that it takes to reduce the concentration level of the dye to !Þ!# grams/gallon. At that time, the amount of dye in the pool is "ß #!! grams. Using the answer in part a b, , we have &!!! / !Þ# Xœ "#!!. Taking the natural logarithm of both sides of the equation results in the required time X œ (Þ"% hours.
a b/. Note that !Þ# œ #!!Î"!!!. Consider the differential equation
.; < .> "!!!
œ ;.
Here the parameter < corresponds to the flow rate , measured in gallons per minute. Using the same initial value, the solution is given by ; >a b œ &!!! / < >^ Î"!!!Þ In order to determine the appropriate flow rate, set > œ % and ; œ "#!!. (Recall that "#!! gm of
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dye has a concentration of !Þ!# gm/gal ). We obtain the equation"#!! œ &!!! / <^ Î#&!Þ Taking the natural logarithm of both sides of the equation results in the required flow rate < œ $&( gallons per minute.
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& > ^2 "! > ^2 68 > % > ^2 68 > œ! ÞHence both functions are solutions of the differential equation.
- C >a b œ a-9= > 68 -9= > > =38 > Ê C b wa b> œ a=38 > 68 -9= > > -9= > b and C wwa b> œ a-9= > 68 -9= > > =38 > =/- > b. Substituting into the left hand side of the differential equation, we have a a -9= > 68 -9= > > =38 > =/- >b b a-9= > 68 -9= > b > =38 > œ a-9= > 68 -9= > > =38 > =/- > b a-9= > 68 -9= > > =38 > œ =/- > b. Hence the function C >a bis a solution of the differential equation.
- Let C >a b œ / <>^. Then C ww^ a b> œ < /#^ <>, and substitution into the differential equation results in < /#^ <>^ # / <>^ œ !. Since / <>^ Á! , we obtain the algebraic equation< # # œ !Þ
The roots of this equation are < (^) "ß# œ „ 3 È# Þ
- C >a b œ / <>^ Ê C w^ a b> œ < / <>^ and C ww^ a b> œ < /#^ <>. Substituting into the differential equation, we have < /#^ <>^ ^ ' / <>^ œ!. Since / <>Á !, we obtain the algebraic equation < #^ < ' œ! , that is, a< # ba< $ b œ!. The roots are < (^) "ß#œ $ , #.
- Let C >a b œ / <>^. Then C w^ a b> œ ^ , C ww^ a b> œ < /#^ <>^ and C wwwa b^ > œ < /$^ <>. Substituting the derivatives into the differential equation, we have < /$^ <>^ $< /#^ <>^ #œ !. Since / <>^ Á! , we obtain the algebraic equation < $^ $< # #< œ! Þ By inspection, it follows that < < "a ba< # b œ!. Clearly, the roots are < (^) " œ! , < (^) # œ " and< (^) $œ # Þ
- C >a b œ > <^ Ê C w^ a b> œ < > <^ "^ and C ww^ a b> œ < < " >a b<#. Substituting the derivatives into the differential equation, we have > #^ c< < " > a b <^ #d^ %> < >a <^ "b % > <œ !. After some algebra, it follows that < < " > %< > % >a b <^ <^ <œ!. For > Á !, we obtain the algebraic equation < #^ &< % œ! Þ The roots of this equation are < (^) " œ " and< (^) #œ % Þ
- The order of the partial differential equation is two , since the highest derivative, in fact each one of the derivatives, is of second order. The equation is linear , since the left hand side is a linear function of the partial derivatives.
- The partial differential equation is fourth order , since the highest derivative, and in fact each of the derivatives, is of order four. The equation is linear , since the left hand side is a linear function of the partial derivatives.
- The partial differential equation is second order , since the highest derivative of the function? Bß Ca b is of order two. The equation is nonlinear , due to the product? † ?Bon the left hand side of the equation. 25.? (^) "aBß C b œ -9= B -9=2 C Ê ?B œ -9= B -9=2 C and?C œ -9= B -9=2 C Þ
(^) " #"
It is evident that ?B ?C Likewise, given #^ #, the second
(^) " #"
^ # œ! Þ^?^ #aBß C^ b^ œ 68 Ba^ Cb
derivatives are
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`? # %B
`B B C
œ B C ? # %CC B C
œ B C
# # (^) #
# # (^) #
2
2
a b
a b
Adding the partial derivatives,
?? # %B # %C BC B C B C
œ B C B C
œ
B C B C
œ!
# #
# # # # # # # #
# (^) #
2 2 a b a b
a b
a B Cb
Hence ?# aBß C bis also a solution of the differential equation.
- Let? (^) "aBß > b œ =38 - B =38 - +>. Then the second derivatives are
?B
œ =38 B =38 +>
?>
œ + =38 B =38 +>
"
"
It is easy to see that + #^ ?B œ ?>. Likewise, given? (^) #Bß > œ =38 B +> , we have
#" #"^ a^ b^ a^ b
?B
œ =38 B +>
?>
œ + =38 B +>
a b
a b
Clearly,? (^) #aBß > bis also a solution of the partial differential equation.
- Given the function? Bß >a b œ È 1 Î> / B Î%^ >, the partial derivatives are
(^) !#
? œ
Î> / Î> B /
? œ
B /
BB
B Î% > # B Î% >
%
>
B Î% >
B Î% >
È È
È È
È
# #
# #^
!!
!!
It follows that !#^? (^) BB œ? œ > #^ % >B> />. È ˆ ‰ È
1! !
# B Î%#^ #>
!
Hence? Bß >a bis a solution of the partial differential equation.
